Translations of known graphs

Shift a graph left, right, up and down by changing its equation, and use translations to find the vertex of a parabola or the centre of a circle.

Shifting left and right

To shift $y = f(x)$ right by $h$, replace $x$ with $x - h$. The new graph is $y = f(x - h)$.

Shifting up and down

To shift $y = f(x)$ up by $k$, add $k$: the new graph is $y = f(x) + k$. Equivalently, replace $y$ with $y - k$.

Translations

$$y = f(x - h) + k \quad\text{is } y = f(x) \text{ shifted right } h \text{ and up } k$$

Translate a graph

Faint: the original graph y = f(x) = x³ − 3x. Orange: the translated graph.

Try this

Set $h = 2$. Notice the equation shows $x - 2$, yet the graph moves right.
Find $h$ and $k$ that move the left turning point $(-1, 2)$ to the origin.

Parabolas and circles

Every parabola $y = a(x - h)^2 + k$ is $y = ax^2$ translated so its vertex moves from $(0, 0)$ to $(h, k)$. Every circle $(x - h)^2 + (y - k)^2 = r^2$ is $x^2 + y^2 = r^2$ translated so its centre moves to $(h, k)$. To find the translation, complete the square.

Example 1●●

Describe $y = x^2 + 4x + 1$ as a translation of $y = x^2$.

$y = (x + 2)^2 - 3$

Half of 4 is 2, and 2² = 4.

Shift left $2$ and down $3$. The vertex moves from $(0, 0)$ to $(-2, -3)$.

Example 2●●

The graph of $y = \dfrac{1}{x}$ is shifted right $3$ and down $2$. Write its new equation and asymptotes.

$y = \dfrac{1}{x - 3} - 2$

Replace x with x − 3, then subtract 2.

Asymptotes $x = 3$ and $y = -2$.

The asymptotes x = 0 and y = 0 move with the graph.

Flashcards

Shift right h

Replace $x$ with $x - h$.

Shift up k

Add $k$ to the function (or replace $y$ with $y - k$).

y = f(x + 3)

$y = f(x)$ shifted left $3$.

Vertex of y = a(x − h)² + k

$(h, k)$

Practice

$y = (x - 5)^2$ is $y = x^2$ shifted:
$y = \sqrt{x}$ shifted left $4$ and up $1$ is:
The circle $(x + 1)^2 + (y - 6)^2 = 4$ is $x^2 + y^2 = 4$ shifted. What is the $x$-coordinate of the new centre?
$y = 2^x$ is shifted down $3$. The new horizontal asymptote is:
Write $y = x^2 - 6x + 5$ as $(x - 3)^2 + k$. Find $k$.
The point $(2, 7)$ is on $y = f(x)$. Which point must be on $y = f(x - 1) + 3$?
$y = \dfrac{1}{x + 2}$ has a vertical asymptote at $x = \ ?$
Which translation takes $x^2 + y^2 = 9$ to $x^2 - 4x + y^2 = 5$?